Question

resolve into partial fractions 2x/(x^2+1)(x-1)​

Answer 1

Given : 2x/(x²+1)(x-1)​

To Find : partial fractions

Solution:

2x/(x²+1)(x-1)​  = (Ax + B)/(x²+1)  + C/(x - 1)

=> 2x = (Ax + B)(x - 1)  + C(x²+1)

x = 1

=> 2(1) = (A(1) + B)(1 - 1) + C(1² + 1)

=> 2 =  (A + B)0  = C(2)

=> 2 = 2C

=> C = 1

2x = (Ax + B)(x - 1)  + 1(x²+1)

x = 0

=> 0  = B (-1)  + 1(1)

=> B = 1

2x = (Ax + 1)(x - 1)  + 1(x²+1)

x   = - 1

=> 2(-1) = (-A + 1)(-1  - 1)  + 1( 1 + 1)

=> - 2 = 2A   - 2  + 2

=> A = - 1

2x = (-x + 1)(x - 1)  + 1(x²+1)

2x/(x²+1)(x-1)​  = (-x + 1)/(x²+1)  + 1/(x - 1)

=> 2x/(x²+1)(x-1)​  = (1 -x)/(x²+1)  + 1/(x - 1)

2x/(x²+1)(x-1)​  = (1 -x)/(x²+1)  + 1/(x - 1)

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